Part CCSIR NET June 2024the-jump-condition-fixes-the-sign

The jump condition fixes the sign

Consider the boundary value problem (BVP) (e^(−5x)y′)′ + 6e^(−5x)y = −f(x), 0 < x < ln 2, with y(0) = 0, y(ln 2) = 0. If $G(x, \xi) = (e^(3x) +$ Be^(2x))(Ce $^(2\xi) +$ De $^(3\xi))$ for $0 \le \xi \le x$ , and $G(x, \xi) = (e^(3\xi) +$ Be $^(2\xi))($ Ce^(2x) + De^(3x)) for $x \le \xi \le \ln 2 ($ Green's function) is such that $\int_{0}^(\ln 2) G(x, \xi)f(\xi)d\xi$ is the solution of the BVP, then the values of B, C and D are

A.B = −2, C = −1, D = 1
B.B = −2, C = 1, D = −1✓ Correct
C.B = 2, C = 1, D = 1
D.B = 2, C = −1, D = −1

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward, and why this one tests execution slip.

See pricing

50 are analysed free — try those first.

The trap it tests

Execution slip

The idea was right. The computation was not.

Drill statements like this

Related counterexample: Every boundary value problem has a Green's function

The jump condition fixes the sign

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